04 Jun 07
Originally posted by agrysonBut they're not equally weighted. That's your mistake.
Stop getting all 'specific example' otherwise you'll keep confusing the question.
The question is... Total average x = 30
I have two averages (y, z), ___equally weighted___, one being unknown, the other being 15.
y = 15
z = ?
(y+z)/2 = x
(15 + z)/2 = 30
=> 15 + z = 2.30
=> z = 2.30 - 15
z = 60 - 15 = 45
This will work for any un ...[text shortened]... iles per hour or pounds of butter per minute, you name it. It's also a generalised solution.
04 Jun 07
I apologise to everyone I thought was wacko. I was wrong, We've done half our journey (distance), our average speed is half of what it needs to be.
To double our average speed, we need to cover three times the distance in the same time, but the question has limited the distance we can travel (D/2)
To double our average speed with the distance remaining we always have exactly 0 time.
This has some interesting possibilities... Pinky, come with me...
04 Jun 07
Originally posted by agrysonSpecific examples are perfectly valid for what we were trying to prove. We were trying to prove that a given generalised answer was wrong. A specific counter-example is all you need to do that.
Stop getting all 'specific example' otherwise you'll keep confusing the question.
04 Jun 07
1 edit
Originally posted by agrysonThey're not equally weighted because the time needed to cover the distance is different.
Stop getting all 'specific example' otherwise you'll keep confusing the question.
The question is... Total average x = 30
I have two averages (y, z), ___equally weighted___, one being unknown, the other being 15.
y = 15
z = ?
(y+z)/2 = x
(15 + z)/2 = 30
=> 15 + z = 2.30
=> z = 2.30 - 15
z = 60 - 15 = 45
This will work for any un ...[text shortened]... iles per hour or pounds of butter per minute, you name it. It's also a generalised solution.
If you travel at 15 mph and you do a mile, you spend for minutes.
If you travel at 45 mph, you'll only be at an average of 30 mph after 4 minutes. Which means you'd need 3 miles to reach average speed of 30 mph.
I think that your mistake is thinking that the average speed of a trip is averaged by distance when, in fact, it is averaged by time as we are averaging speed over distance.
04 Jun 07
Originally posted by agrysonMy last post is what happens when you reply before reading a thread to the end...
I apologise to everyone I thought was wacko. I was wrong, We've done half our journey (distance), our average speed is half of what it needs to be.
To double our average speed, we need to cover three times the distance in the same time, but the question has limited the distance we can travel (D/2)
To double our average speed with the distance remaining w ...[text shortened]... lways have exactly 0 time.
This has some interesting possibilities... Pinky, come with me...
04 Jun 07
Originally posted by abejnoodIf everyone just reads this post the reasoning behind the answer will become clear.
Which is why you have to go at infinite speed! We agree that speed or velocity = displacement divided by time. So you have to double your velocity, right?
Assume the distance is 2 miles. You go one mile at 15 mph. So it takes four minutes. Now, in order for the average speed for the trip to be 30 miles an hour, the truck has to travel one mile every to ...[text shortened]... That's why it needs to travel infinitely fast: it has to go a mile with no time passing at all.
04 Jun 07
If everyone just reads this post the reasoning behind the answer will become clear.Yes it is clear. You are insane abejnood. 😉 You have made a measurement, in this case an object moving at an average speed of 30 mph. 🙄 It has to have travelled a certain distance for this measurement to have been made. 🙄 You have stated yourself that is has only moved 1/2 of a certain distance, the other 1/2 still remaining. 🙄 It still has to travel the other half of it's journey. 🙄 You cannot increase it's speed or change the measurement. 🙄 Clearly you don't understand simple science or simple scientific principles.🙄
04 Jun 07
2 edits
They're not equally weighted because the time needed to cover the distance is different.I think you have lost your marbles Palynka. 🙄 An average measurement is an average of two values, they are not allowed to shift or change. If you could expand on this and provide some real world examples, instead of making these dull-minded assumption and tossing more absurd and whack definitions around, i'd be much obliged.The changing flexible time thing is contradicted by the problem definition, mainly that it has travelled 1/2 of a unspecified distance. And rewording the problem to suit your silly answer is rather silly.
I think that your mistake is thinking that the average speed of a trip is averaged by distance when, in fact, it is averaged by time as we are averaging speed [b]over distance.[/b]